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Q.

A particle performing S.H.M. is found at its equilibrium at t= 1s and it is found to have a speed of 0.25 m/s at t=2s. If the period of oscillation is 6s. Calculate amplitude of oscillation.

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a

38πm

b

34πm

c

6πm

d

32πm

answer is A.

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Detailed Solution

x=Asinωt+ϕ   At t=1 second    x=0       0= Asinω×1+ϕϕ=-ω Now   V=Aωcosωt+ϕ At t=1 second    V=14m/s So    14=A×ωcosω×2-ω         14=Aωcosω

by putting values

14=A×2π6cos2π6     so A=32πm

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